/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


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YES
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished


Termination w.r.t. Q of the given QTRS could be proven:

(0) QTRS
(1) QTRSRRRProof [EQUIVALENT, 70 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 4 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) QDP
(7) QDPOrderProof [EQUIVALENT, 53 ms]
(8) QDP
(9) PisEmptyProof [EQUIVALENT, 0 ms]
(10) YES


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(0)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   a(x1) -> b(b(x1))
   a(b(b(x1))) -> b(b(c(c(c(a(x1))))))
   b(b(x1)) -> c(c(c(x1)))
   c(c(c(b(b(x1))))) -> a(x1)

Q is empty.

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(1) QTRSRRRProof (EQUIVALENT)
Used ordering:
Polynomial interpretation [POLO]:

   POL(a(x_1)) = 2 + x_1
   POL(b(x_1)) = 1 + x_1
   POL(c(x_1)) = x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   b(b(x1)) -> c(c(c(x1)))




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(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   a(x1) -> b(b(x1))
   a(b(b(x1))) -> b(b(c(c(c(a(x1))))))
   c(c(c(b(b(x1))))) -> a(x1)

Q is empty.

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(3) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A(b(b(x1))) -> C(c(c(a(x1))))
   A(b(b(x1))) -> C(c(a(x1)))
   A(b(b(x1))) -> C(a(x1))
   A(b(b(x1))) -> A(x1)
   C(c(c(b(b(x1))))) -> A(x1)

The TRS R consists of the following rules:

   a(x1) -> b(b(x1))
   a(b(b(x1))) -> b(b(c(c(c(a(x1))))))
   c(c(c(b(b(x1))))) -> a(x1)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.
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(6)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   C(c(c(b(b(x1))))) -> A(x1)
   A(b(b(x1))) -> C(c(c(a(x1))))
   A(b(b(x1))) -> A(x1)

The TRS R consists of the following rules:

   a(x1) -> b(b(x1))
   a(b(b(x1))) -> b(b(c(c(c(a(x1))))))
   c(c(c(b(b(x1))))) -> a(x1)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(7) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   C(c(c(b(b(x1))))) -> A(x1)
   A(b(b(x1))) -> C(c(c(a(x1))))
   A(b(b(x1))) -> A(x1)
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial interpretation [POLO]:

   POL(A(x_1)) = 4 + 4*x_1
   POL(C(x_1)) = 2 + 4*x_1
   POL(a(x_1)) = 3 + 4*x_1
   POL(b(x_1)) = 1 + 2*x_1
   POL(c(x_1)) = x_1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   a(x1) -> b(b(x1))
   a(b(b(x1))) -> b(b(c(c(c(a(x1))))))
   c(c(c(b(b(x1))))) -> a(x1)


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(8)
Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:

   a(x1) -> b(b(x1))
   a(b(b(x1))) -> b(b(c(c(c(a(x1))))))
   c(c(c(b(b(x1))))) -> a(x1)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(9) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
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(10)
YES